SRM 392(1-250pt)

DIV1 250pt

题意：给两个各含有一个*号的字符串s1和s2，可以用一个任意字符串代替*号(注意是串，不是只能用单个字符代替，也可以为用空串代替)，问能否将s1和s2变为相同的字符串。如果能输出改变后长度最短的方案。

解法：其实是很简单的暴力。。。。我代码写的慢有两个原因，一是string的substr不会用，另一个是因为s1和s2地位对等，可以通过交换他们的位置来省代码。

tag:brute-force

// BEGIN CUT HERE
/*
 * Author:  plum rain
 * score :
 */
/*

 */
// END CUT HERE
#line 11 "TwoStringMasks.cpp"
#include <sstream>
#include <stdexcept>
#include <functional>
#include <iomanip>
#include <numeric>
#include <fstream>
#include <cctype>
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <queue>
#include <bitset>
#include <list>
#include <string>
#include <utility>
#include <map>
#include <ctime>
#include <stack>

using namespace std;

#define clr0(x) memset(x, 0, sizeof(x))
#define clr1(x) memset(x, -1, sizeof(x))
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false)

typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<double> vd;
typedef pair<int, int> pii;
typedef long long int64;

const double eps = 1e-8;
const double PI = atan(1.0)*4;
const int inf = 2139062143 / 2;

string temp = "impossible";

string gao(string s1, string s2)
{
    int len = sz(s1);
    string ans;
    for (int i = len; i; -- i){
        int t1 = 0, t2 = sz(s2) - i;
        bool ok = 1;
        while (t2 < sz(s2) && t1 < len){
            if (s1[t1] != s2[t2]){
                ok = 0; break;
            }
            ++ t1; ++ t2;
        }
        if (ok){
            for (int j = i; j < len; ++ j)
                s2.pb (s1[j]);
            return s2;
        }
    }
    return s2 + s1;
}

string Out(string s)
{
    string ans;
    for (int i = 0; i < sz(s); ++ i)
        if (s[i] != '*') ans.pb (s[i]);
    return ans;
}

class TwoStringMasks
{
    public:
        string shortestCommon(string s1, string s2){
            int i1 = 0, i2 = sz(s1)-1, j1 = 0, j2 = sz(s2)-1;
            while (s1[i1] != '*' && s2[j1] != '*'){
                if (s1[i1] == s2[j1]) ++ i1, ++ j1;
                else return temp;
            }
            while (s1[i2] != '*' && s2[j2] != '*'){
                if (s1[i2] == s2[j2]) -- i2, -- j2;
                else return temp;
            }

            if (s1[i1] == s1[i2] && s1[i1] == '*') return Out(s2);
            if (s2[j1] == s2[j2] && s2[j1] == '*') return Out(s1);
            
            if (s1[i1] == '*' && s1[i2] != '*'){
                string t1, t2, g1, g2;
                for (int i = 0; i < i1; ++ i) t1.pb (s1[i]);
                for (int i = i2+1; i < sz(s1); ++ i) t2.pb (s1[i]);
                for (int i = i1+1; i <= i2; ++ i) g1.pb (s1[i]);
                for (int i = j1; i < j2; ++ i) g2.pb (s2[i]);
                return t1 + gao(g1, g2) + t2;
            }
            if (s1[i1] != '*' && s1[i2] == '*'){
                string t1, t2, g1, g2;
                for (int i = 0; i < i1; ++ i) t1.pb (s1[i]);
                for (int i = i2+1; i < sz(s1); ++ i) t2.pb (s1[i]);
                for (int i = i1; i < i2; ++ i) g1.pb (s1[i]);
                for (int i = j1+1; i <= j2; ++ i) g2.pb (s2[i]);
                return t1 + gao(g2, g1) + t2;
            }
            return temp;
        }
        
// BEGIN CUT HERE
    public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) test_case_5(); if ((Case == -1) || (Case == 6)) test_case_6(); if ((Case == -1) || (Case == 7)) test_case_7(); }
    //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_6();}
    private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
    void verify_case(int Case, const string &Expected, const string &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
    void test_case_0() { string Arg0 = "TOPC*DER"; string Arg1 = "T*PCODER"; string Arg2 = "TOPCODER"; verify_case(0, Arg2, shortestCommon(Arg0, Arg1)); }
    void test_case_1() { string Arg0 = "HELLO*"; string Arg1 = "HI*"; string Arg2 = "impossible"; verify_case(1, Arg2, shortestCommon(Arg0, Arg1)); }
    void test_case_2() { string Arg0 = "GOOD*LUCK"; string Arg1 = "*"; string Arg2 = "GOODLUCK"; verify_case(2, Arg2, shortestCommon(Arg0, Arg1)); }
    void test_case_3() { string Arg0 = "*SAMPLETEST"; string Arg1 = "THIRDSAMPLE*"; string Arg2 = "THIRDSAMPLETEST"; verify_case(3, Arg2, shortestCommon(Arg0, Arg1)); }
    void test_case_4() { string Arg0 = "*TOP"; string Arg1 = "*CODER"; string Arg2 = "impossible"; verify_case(4, Arg2, shortestCommon(Arg0, Arg1)); }
    void test_case_5() { string Arg0 = "*"; string Arg1 = "A*"; string Arg2 = "A"; verify_case(5, Arg2, shortestCommon(Arg0, Arg1)); }
    void test_case_6() { string Arg0 = "*A"; string Arg1 = "B*"; string Arg2 = "BA"; verify_case(6, Arg2, shortestCommon(Arg0, Arg1)); }
    void test_case_7() { string Arg0 = "LASTCASE*"; string Arg1 = "*LASTCASE"; string Arg2 = "LASTCASE"; verify_case(7, Arg2, shortestCommon(Arg0, Arg1)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );    
    TwoStringMasks ___test;
    ___test.run_test(-1);
       return 0;
}
// END CUT HERE